删除 2 个字符串的字符，使它们成为打字稿中的字谜

Question

我试图解决其中一项 Hackerrank 挑战 String: Making Anagrams

我有 2 个这样的字符串：

let a: string = "fcrxzwscanmligyxyvym"
let b: string = "jxwtrhvujlmrpdoqbisbwhmgpmeoke"

我有一个函数只通过了 3 次测试！！！ :

function makeAnagram(a: string, b: string): number {
    type Map = {
        [key :string] : number
    }
   let string1 = a.split('')
   let string2 = b.split('')
   let map : Map = {}
   let deletedCount = 0
   let firstCount = 0
   let isMoreThanTwo = false
   
   for(let i of string1) {
       map[i] = (map[i] | 0) + 1
   }
   
   for(let i of string2){
       map[i] = (map[i] | 0) + 1
   }
   
   for(let i in map) {
       if(map[i] == 1) {
           deletedCount++
           firstCount++
       } else if(map[i] > 2) {
           isMoreThanTwo = true
           deletedCount += (map[i] - 1)
       }
       
   }
   
   return isMoreThanTwo ? deletedCount + firstCount : deletedCount

有没有其他方法可以统计删除的字符数？你们能给我一些建议吗，谢谢

Answer 1

我刚刚解决了这个问题，但在通过所有测试用例之前，我找到了我使用的解决方案

for (let [k, v] of map) {
  remain += v;
}

失败了 13/16 个测试用例，然后我调试了程序，发现有时当我们从这一步的前面减去 1 时，结果是 less than 0 所以哟也必须处理这种情况，我处理为

for (let [k, v] of map) {
  remain += v < 0 ? v * -1 : v;
}

Now, all test cases are passed

function makeAnagram(a, b) {
  const [small, big] = a.length < b.length ? [a, b] : [b, a];
  const map = new Map();
  for (let c of small) {
    map.set(c, (map.get(c) ?? 0) + 1);
  }

  let remain = 0;
  for (let c of big) {
    !map.has(c) ? remain++ : map.set(c, map.get(c) - 1);
  }

  for (let [k, v] of map) {
    remain += v < 0 ? v * -1 : v;
  }

  return remain;
}

console.log(makeAnagram("fast", "sofasty"));

删除 2 个字符串的字符，使它们成为打字稿中的字谜

delete character of 2 string to make them anagram in typescript

javascript

typescript

anagram